*有连续编号为1...N的N个房间，每个房间关押一个犯人，有M种宗教，每个犯人可能信仰其中一种。如果
相邻房间的犯人的宗教相同，就可能发生越狱，求有多少种状态可能发生越狱

　　输入两个整数M，N.1<=M<=10^8,1<=N<=10^12

　　可能越狱的状态数，模100003取余

 //It is made by Awson on 2018.1.13

 #include <set>

 #include <map>

 #include <cmath>

 #include <ctime>

 #include <queue>

 #include <stack>

 #include <cstdio>

 #include <string>

 #include <vector>

 #include <cstdlib>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #define LL long long

 #define Max(a, b) ((a) > (b) ? (a) : (b))

 #define Min(a, b) ((a) < (b) ? (a) : (b))

 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

 using namespace std;

 const int MOD = ;

 void read(LL &x) {

     char ch; bool flag = ;

     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || ); ch = getchar());

     for (x = ; isdigit(ch); x = (x<<)+(x<<)+ch-, ch = getchar());

     x *= -*flag;

 }

 void write(LL x) {

     if (x > ) write(x/);

     putchar(x%+);

 }

 LL n, m;

 LL quick_pow(LL a, LL b) {

     b %= MOD-; a %= MOD; LL ans = ;

     while (b) {

     if (b&) ans = ans*a%MOD;

     a = a*a%MOD, b >>= ;

     }

     return ans;

 }

 void work() {

     read(m), read(n);

     write((quick_pow(m, n)-m*quick_pow(m-, n-)%MOD+MOD)%MOD);

 }

 int main() {

     work();

     return ;

 }